# Why does the minimum value of $x^x$ equal $1/e$?

The graph of $y=x^x$ looks like this:

As we can see, the graph has a minimum value at a turning point. According to WolframAlpha, this point is at $x=1/e$.

I know that $e$ is the number for exponential growth and $\frac{d}{dx}e^x=e^x$, but these ideas seem unrelated to the fact that the mimum value of $x^x$ is $1/e$. Is this just pure coincidence, or could someone provide an intuitive explanation (i.e. more than just a proof) of why this is?

• I would appeal to the logarithmic derivative of the function to show that. – Triatticus Jan 4 '16 at 20:25

Hint: Note that $x^x = e^{x\log x}$.

So minimizing $x^x$ is the same as minimizing $x\log x$

Let $f(x)=x^x$. Note that $f(x)$ is only defined for $x>0$.

Then $$\ln f(x)=x\cdot\ln x\tag{1}$$ Differentiating (1) gives $$\frac{1}{f(x)}f^\prime(x)=x\frac{1}{x}+\ln x=1+\ln x$$ Note that we have used the chain rule and the product rule.

Solving for $f^\prime(x)$ gives $$f^\prime(x)=f(x)(1+\ln x)=x^x(1+\ln x)$$ Can you use this to locate the critical points of $f(x)$?

• I think you can leave the absolute value off since $x>0$ on its domain, right? – Alejandro Jan 4 '16 at 20:31
• @Subjunctive sure – Brian Fitzpatrick Jan 4 '16 at 20:31

Hint: actually you are looking for a local/global minimum .. so look at the derivative of the function $f(x) = x^x$ $$f'(x) = x^x (\log (x)+1)$$ which equals $0 \iff x = \frac{1}{e}$

Try to minimize logarithm of $y= x^x,$ i.e., $y=x \, \log x$

Its derivative is

$$1 + \log(x)$$

When equated to zero, it solves to that the minimum of $y(x)$ ocuurs at

$$x= \dfrac{1}{e}= 0.36788$$

and the minimum value is

$$y_{min}= \dfrac{1}{{e}^{\frac{1}{e}}} \approx 0.6922$$

The tiny red dot shown in your graph for minimum is:

$$(x,y) = (0.36788, 0.6922)$$

Note that the minimum value is not $\dfrac{1}{e}\, ! \,$ ... but is the reciprocal of $e^{th}$root of $e.$